Newcomb's Problem as an Iterated Prisoner's Dilemma

post by Jerdle (daniel-amdurer) · 2022-01-05T22:48:57.661Z · LW · GW · 11 comments

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The intuitively paradoxical aspect of Newcomb's Problem is that it contains a loop. Your decision determines Omega's prediction, which determines the options you decide between. There are many decision theories that solve this problem, usually leading to one-boxing. However, these decision theories are less than intuitive because of their acausal nature. This is a hopefully more intuitive analysis of Newcomb's Problem that unrolls the loop and reveals the Iterated Prisoner's Dilemma within.

First, unroll the loop. To do this, have Omega run purely on past data. To keep the high quality of the prediction, the past data must also determine your decision at that point. This is easiest to implement with a bot (in the sense of PrudentBot, but much simpler). Specifically, you choose your decision, and that sets up a bot that only performs that decision. After unrolling, we now have an iterated game, as the dependence on the same round has been replaced with one on a previous round.

Now, from your perspective, your income is as follows.

	Fill B	Leave B empty
Take A and B	$1001000	$1000
Take B	$1000000	$0

Now, relabelling.

	Cooperate	Defect
Cooperate	$1000000	$0
Defect	$1001000	$1000

This is just the Prisoner's Dilemma! And as it was iterated by unrolling, it is the Iterated Prisoner's Dilemma.

Omega plays TitForTat, while you can choose between CooperateBot and DefectBot. As CooperateBot performs better against TitForTat than DefectBot, you're best off cooperating. As cooperating corresponds to one-boxing, you should one-box.

Rerolling the loop, you should one-box in the original problem.

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